Name: An Equation for Velocity and Acceleration in SHM
Uploaded: 2019-07-19
Duration: 14 min 10 s
Description: Velocity and acceleration in SHM are the foundations of many concepts in physics. Watch this video to find equations for the velocity and acceleration of a p...

Question

Derive the expressions for displacement, velocity and acceleration of a particle executing S.H.M.

Accepted Answer

To derive the expression for displacement of the particle, let's consider a particle P moves on the circumference of circle of radius $A$ with uniform acceleration. Let PN is the perpendicular drawn from P to diameter $y y'$ .

From the given figure, $∠ P O X = θ$ , $O P = A$ and $O N = y$
From the right-angled triangle $∆ O N P$ ,

$\sin θ = \frac{O N}{O P}$

Since, $θ = ω t$

$\Rightarrow O N = O P \sin (ω t)$

Substituting the values from above,

$y = A \sin (ω t)$ ----(1)

This is the expression of equation of SHM.

Since $v = \frac{d y}{d t}$
Differentiate equation (1) with respect to time $(t)$ , to find velocity of particle.

$v = \frac{d (A \sin (ω t))}{d t}$

$v = A ω \cos (ω t)$ .....(2)

From the figure, $\cos (ω t) = \frac{\sqrt{A^{2} - y^{2}}}{A}$

$\Rightarrow v = ω \sqrt{A^{2} - y^{2}}$

This is the expression of velocity of particle in SHM.

Acceleration of the particle $a = \frac{d v}{d t}$

Differentiate equation (2) with respect to time $(t)$ ,

$a = \frac{d (A ω \cos (ω t))}{d t}$

$a = - ω^{2} A \sin (ω t)$

From equation (1),

$\Rightarrow a = - ω^{2} y$
This is the expression of acceleration of particle in SHM.

Derive the expressions for displacement, velocity and acceleration of a particle executing S.H.M.

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